What is (\partial)/(\partial y)(((ye^x))/(xe^y)) ?

The answer to (\partial)/(\partial y)(((ye^x))/(xe^y)) is (e^{x-y}(1-y))/x

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(\partial)/(\partial y)(((ye^x))/(xe^y))

\frac{\partial}{\partial y} (\frac{( y e ^{x} )}{x e ^{y}})

\frac{e ^{x - y} ( 1 - y )}{x}

Solution steps

\frac{\partial}{\partial y} (\frac{( y e ^{x} )}{x e ^{y}})

Treat

x

as a constant

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= \frac{e ^{x}}{x} \frac{\partial}{\partial y} (\frac{y}{e ^{y}})

Apply the Quotient Rule:

(\frac{f}{g})^{'} = \frac{f ^{'} \cdot g - g ^{'} \cdot f}{g ^{2}}

= \frac{e ^{x}}{x} \cdot \frac{\frac{\partial}{\partial y} ( y ) e ^{y} - \frac{\partial}{\partial y} ( e ^{y} ) y}{( e ^{y} ) ^{2}}

\frac{\partial}{\partial y} (y) = 1

\frac{\partial}{\partial y} (e^{y}) = e^{y}

= \frac{e ^{x}}{x} \cdot \frac{1 \cdot e ^{y} - e ^{y} y}{( e ^{y} ) ^{2}}

Simplify

\frac{e ^{x}}{x} \cdot \frac{1 \cdot e ^{y} - e ^{y} y}{( e ^{y} ) ^{2}} : \frac{e ^{x - y} ( 1 - y )}{x}

= \frac{e ^{x - y} ( 1 - y )}{x}

f (x) = e^{x} - e^{- x}

f (t) = (t + 1)^{3}

(- 1, - 1), (1, - 4)

\int x^{2} sin (θ x) d x

\int_{0}^{5} π (x - 1)^{2} d x

What is (\partial)/(\partial y)(((ye^x))/(xe^y)) ?
The answer to (\partial)/(\partial y)(((ye^x))/(xe^y)) is (e^{x-y}(1-y))/x